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emcrazy14

  • 3 years ago

Differentiate; x^2/(1+x)^2

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  1. ajprincess
    • 3 years ago
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    This is of the form u/v. (u/v)'=(u'v-v'u)/v^2 Here u=x^2 and v=(1+x)^2. (x^n)'=nx^(n-1) Can u do it @emcrazy14?

  2. emcrazy14
    • 3 years ago
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    I got the answer -2x/(1+x)^3. Is it correct?

  3. experimentX
    • 3 years ago
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    differentiate: 1/(1/x + 1)^2

  4. ajprincess
    • 3 years ago
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    check the sign @emcrazy14. It shud be 2x/(1+x)^3

  5. emcrazy14
    • 3 years ago
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    Ok, thanks. Do I multiply (x+1) with a negative sign to change it into (1+x) so that it cancels with the the denominator (1+x)^4?

  6. experimentX
    • 3 years ago
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    just ask the wolf for differentiation Question.

  7. ajprincess
    • 3 years ago
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    nope. (x+1)=(1+x)

  8. emcrazy14
    • 3 years ago
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    Then how oes it become +2x? :O

  9. emcrazy14
    • 3 years ago
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    @experimentX : wolf? :O *does

  10. experimentX
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=differentiate+x^2%2F%281%2Bx%29^2

  11. ajprincess
    • 3 years ago
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    |dw:1352469822062:dw||dw:1352469940166:dw||dw:1352470079425:dw|

  12. experimentX
    • 3 years ago
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    \[ x^2/(1+x)^2 = 1/(1/x + 1)^2 = ({1 \over x} + 1)^{-2}\] use chain rule ... this should be shorter.

  13. emcrazy14
    • 3 years ago
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    @ajprincess : Get it now. Thank you so much! :D

  14. ajprincess
    • 3 years ago
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    welcome:) @experimentX's method is much more simpler. U can use that too.:)

  15. emcrazy14
    • 3 years ago
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    @experimentX : Once I perfect this one, I'll use the one you mentioned. Thank you! :)

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